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Find $N$ so that for all $n\geq N$ holds that : $$ 2^{-n} + 3^{-n} + 4^{-n} < \frac{1}{365} $$

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Finding some $N$ is easy... do you actually want the smallest $N$? And is this homework? – Alastair Litterick Apr 16 '12 at 9:32
$N=9$ by trial and error... not very enlightening... – bgins Apr 16 '12 at 9:47

A little playing around might help... $$ 2^{-n}+3^{-n}+4^{-n}<\frac1{365} $$ $$ 6^n+4^n+3^n<\frac{12^n}{365} $$ $$ 1+\left(\frac23\right)^n+\left(\frac12\right)^n<\frac{2^n}{365} $$ The LHS decreases from $3$ to $1$ for $n\in\mathbb{N}$, while the RHS increases from $\frac1{365}$ to $\infty$. Graphing or trial and error soon finds the meeting point, which over (by rootfinding) $\mathbb{R}$ is $8.55963662579$. Therefore, $N=9$.

Even without rootfinding, we can answer this since $2^8=256$ so that the inequality is clearly false for $n=8$, while $2^9=512$ so that for $n=9$, the RHS becomes $$\frac{512}{365}=1+\frac{147}{365}>1+\frac{512}{19683}+\frac{1}{512},$$ and without resorting to such high powers, one can note that $$\left(\frac23\right)^2=\frac49<\frac12$$ so that the LHS satisfies $$ 1+\left(\frac23\right)^n+\left(\frac12\right)^n < 1+\left(\frac12\right)^{n/2}+\left(\frac12\right)^n $$ and conclude $$ 1+\left(\frac23\right)^9+\left(\frac12\right)^9 < 1+\left(\frac12\right)^4+\left(\frac12\right)^9 = 1+\frac1{16}+\frac1{512} < 1+\frac1{10} < \frac{512}{365} \,. $$

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Note that if $n >0$ then $\dfrac{1}{3^n}<\dfrac{1}{2^n}$ and $\dfrac{1}{4^n}<\dfrac{1}{2^n}$. So $$\frac{1}{2^n}+\frac{1}{3^n}+\frac{1}{4^n}<\frac{3}{2^n}.$$

Thus if $\dfrac{3}{2^n}<\dfrac{1}{365}$, the desired inequality will hold. So let's make sure that $\dfrac{1}{2^n}<\dfrac{1}{3\cdot 365}$, or equivalently that $2^n>1095$. We are familiar with powers of $2$: certainly $2^{11}=2048>1095$. Therefore we can take $N=11$.

Remark: There is a smaller $N$ that will work. But the question did not ask for the cheapest $N$ that works. Sometimes, in doing estimates to prove convergence, students expend excessive effort finding the smallest $N$ such that $|a_n-a|<\epsilon$ for all $n>N$.

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I thank you for your answer. The question came on the theory in seauences, where we had to solve limits of sequences. This execercise on series undermined the theory completely. – Ignace Apr 16 '12 at 14:19

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